Find the HCF of 36, 108, and 156.

Correct option is A

Given:

The numbers are 36, 108, and 156.

Formula Used:

HCF(a,b,c) = Product of common prime factors of all three numbers ​​

Solution:

Prime factorization:

36 = $$2^2 \times 3^2$$​​

108 = $$2^2 \times 3^3$$​​

156 = $$2^2 \times 3 \times 13$$​​

Common factors in all three:

​$$2^2$$ is common​

3 is common (minimum power = 1)

​$$\text{HCF} = 2^2 \times 3 = 4 \times 3 = {12}$$​​

Thus, the correct option is (a) 12